An ideal gas of volume 2 ft cubed is contained within a container having an area

Question

An ideal gas of volume 2 ft cubed is contained within a container having an area of 1 ft squared and a volume of 5 ft cubed. The pressure of the gas is 10 atm, the force of the piston is 1440 lbf, and an external force is applied to keep the piston in place. The external pressure is 1 atm. When the external force is released, the piston rises without friction up the container. The gas in the container is always kept at the same temperature. What is the velocity of the moving piston?

Hint: for ideal gas. PV=nRT PV=constant An ideal gas of volume 2 ft cubed is contained within a container having an area of 1 ft squared and a volume of 5 ft cubed. The pressure of the gas is 10 atm, the force of the piston is 1440 lbf, and an external force is applied to keep the piston in place. The external pressure is 1 atm. When the external force is released, the piston rises without friction up the container. The gas in the container is always kept at the same temperature. What is the velocity of the moving piston?

Hint: for ideal gas. PV=nRT PV=constant

Hint: for ideal gas. PV=nRT PV=constant

Explanation / Answer

As the temperature T is constant, it is a isothermal expansion. With Isothermal process work done

W = nRTln(V2/V1)

Initial volume = 2cft, final volume = 5 cft

W = nRTln(5/2)

With initial condition

10*14.696*144*2 = nRT   ( 1 atm = 14.696 psi)

W = 10*14.696*2*ln(5/2)

         = 38781.528

If v is the velocity of the piston and m is the mass then

mv2 /2 = 38781.528

mass of the piston = 1440/32.17 = 44.76 lbs

average velocity of the piston = sqrt(38781.53*2/44.76)

                                                        = 41.63 ft/s

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